What irrational number can be added to pi to get a sum that is rational

Mathematics

1 answer:

m_a_m_a [10]3 years ago

5 0

Answer is -pi

Let x be this unknown number. Adding on x to pi gets us the expression x+pi

If we set x+pi equal to zero (which is a rational number) and solve for x, we get

x+pi = 0

x+pi - pi = 0 - pi

x = -pi

so adding pi to -pi will lead to 0 since

x+pi = -pi+pi = 0

You might be interested in

What is the square root of 25? =7

nataly862011 [7]

Square root of 25 is 5

4 0

2 years ago

Read 2 more answers

What is 26.73 rounded to the nearest decimal

Goshia [24]

I think it's 26.70 if I remember what a decimal is.

5 0

3 years ago

A sequence is defined by the recursive formula f(n + 1) = 1.5f(n). Which sequence could be generated using the formula? –12, –18

AlexFokin [52]

It is -16, 17.5, -19, ...

4 0

3 years ago

Independent Practice

Sedbober [7]

Answer: THE ANSWER IS D. (-27, -7, -2, 8, 48)

Step-by-step explanation:

Graph.

y = 5 x − 2

{ − 5 , − 1 , 0 , 2 , 10 }

The chosen topic is not meant for use with this type of problem. Try the examples below.

1 , 2 , 2 , 2 , 3 , 4 , 4 , 5 , 8 , 9 , 10

11 , 14 , 14 , 17 , 17 , 17 , 41 , 44 , 47 , 71 , 74 , 77

12 , 15 , 45 , 65 , 78

5 0

3 years ago

How many even 5-digit numbers can be formed from the numbers 1, 2, 4, 7, 8 if no digit is repeated in a number? A. 72 B. 120 C.

katovenus [111]

Hello,

Let's place the last digit: it must be 2 or 4 or 8 (3 possibilities)

It remainds 4 digits and the number of permutations fo 4 numbers is 4!=4*3*2*1=24

Thus there are 3*24=72 possibilities.

Answer A

If you do'nt believe run this programm

DIM n(5) AS INTEGER, i1 AS INTEGER, i2 AS INTEGER, i3 AS INTEGER, i4 AS INTEGER, i5 AS INTEGER, nb AS LONG, tot AS LONG
tot = 0
n(1) = 1
n(2) = 2
n(3) = 4
n(4) = 7
n(5) = 8
FOR i1 = 1 TO 5
    FOR i2 = 1 TO 5
        IF i2 <> i1 THEN
            FOR i3 = 1 TO 5
                IF i3 <> i2 AND i3 <> i1 THEN
                    FOR i4 = 1 TO 5
                        IF i4 <> i3 AND i4 <> i2 AND i4 <> i1 THEN
                            FOR i5 = 1 TO 5
                                IF i5 <> i4 AND i5 <> i3 AND i5 <> i2 AND i5 <> i1 THEN
                                    nb = ((((n(i1) * 10) + n(i2)) * 10 + n(i3)) * 10 + n(i4)) * 10 + n(i5)
                                    IF nb MOD 2 = 0 THEN
                                        tot = tot + 1
                                    END IF
                                END IF
                            NEXT i5
                        END IF
                    NEXT i4
                END IF
            NEXT i3
        END IF
    NEXT i2
NEXT i1
PRINT "tot="; tot
END

7 0

3 years ago